\(3.x^3=-\frac{1}{9}\)
\(x^3=-\frac{1}{9}:3\)
\(x^3=-\frac{1}{27}\)
\(x^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x=\frac{-1}{3}\)
b)\(\left(x+1\right)^2=\frac{2}{75}:\frac{2}{3}\)
\(\left(x+1\right)^2=\frac{2}{75}.\frac{3}{2}\)
\(\left(x+1\right)^2=\frac{1}{25}\)
\(\left(x+1\right)^2=\left(\frac{1}{5}\right)^2\)
\(\Rightarrow x+1\in\left\{\frac{1}{5};\frac{-1}{5}\right\}\)
Th1: \(x+1=\frac{1}{5}\)
\(x=\frac{1}{5}-1\)
\(x=\frac{-4}{5}\)
Th2:\(x+1=\frac{-1}{5}\)
\(x=\frac{-1}{5}-1\)
\(x=\frac{-6}{5}\)
Vậy \(x\in\left\{\frac{-4}{5};\frac{-6}{5}\right\}\)
c) \(\left(2x-1\right)^7=\left(3x-1\right)^7\)
\(\Rightarrow2x-1=3x-1\)
\(\Rightarrow2x-3x=1+1\)
\(\Rightarrow x\left(2-3\right)=2\)
\(\Rightarrow x.\left(-1\right)=2\)
\(\Rightarrow x=-2\)
d) \(2^{x+1}.3y=12\)
\(\Rightarrow2^x.2.3y=12\)
\(\Rightarrow2^x.3y=6\)
\(\Rightarrow2^x.3y=1.6=6.1=\left(-1\right)\left(-6\right)=\left(-6\right)\left(-1\right)=2.3=3.2=\left(-2\right).\left(-3\right)=\left(-3\right)\left(-2\right)\)
Ta có bảng:
| 2x | 1 | 6 | -1 | -6 | 2 | 3 | -2 | -3 |
| 3y | 6 | 1 | -6 | -1 | 3 | 2 | -3 | -2 |
| x | 0 | loại | loại | loại | 1 | loại | loại | loại |
| y | 2 | \(\frac{1}{3}\) | -2 | \(\frac{-1}{3}\) | 1 | \(\frac{2}{3}\) | -1 | \(\frac{-2}{3}\) |
Vậy có 2 cặp (x;y) thỏa mãn (0;2) và (1;1)
a, \(3x^3=-\frac{1}{9}\)
\(x^3=-\frac{1}{9}:3=-\frac{1}{27}\)
Vì: \(\left(-\frac{1}{3}\right)^3=-\frac{1}{27}\)
=> x = -1/3
